The drawing shows a skateboarder moving at 6.00 m/s along a horizontal section o

Question

The drawing shows a skateboarder moving at 6.00 m/s along a horizontal section of a track that is slanted upward by = 48.0° above the horizontal at its end, which is 0.650 m above the ground. When she leaves the track, she follows the characteristic path of projectile motion. Ignoring friction and air resistance, find the maximum height H to which she rises above the end of the track.

image ------->https://d2vlcm61l7u1fs.cloudfront.net/media%2F02e%2F02e57194-2674-4d44-ba5f-0d8fbe5c01b4%2FphpKyY5me.png instead of 5.10 m/s it is 6 m/s and instead of .69 m it is .65 m

Explanation / Answer

by the conservation of energy

initial energy = final energy

0.5 * mv^2 = mgh

0.5 * v^2 = gh

0.5 * 6^2 = 9.8 * h

h = 1.836 m

height above the track = 1.836 - 0.65

height above the track = 1.186 m

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